Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles[/tex]
[tex]\text{Moles of} CO=\frac{260}{28}=9.3moles[/tex]
[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)[/tex]
According to stoichiometry :
1 mole of [tex]Fe_2O_3[/tex] require 3 moles of [tex]CO[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] will require=[tex]\frac{3}{1}\times 2.8=8.4moles[/tex] of [tex]CO[/tex]
Thus [tex]Fe_2O_3[/tex] is the limiting reagent as it limits the formation of product and [tex]CO[/tex] is the excess reagent.
As 1 mole of [tex]Fe_2O_3[/tex] give = 2 moles of [tex]Fe[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] give =[tex]\frac{2}{1}\times 2.8=5.6moles[/tex] of [tex]Fe[/tex]
Mass of [tex]Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g[/tex]
Theoretical yield of liquid iron = 313.6 g
Experimental yield = 288 g
Now we have to calculate the percent yield
[tex]\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%[/tex]
Therefore, the percent yield is, 91.8%
A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.
Answer:
19.07 g mol^-1
Explanation:
The computation of the molecular mass of the unknown gas is shown below:
As we know that
[tex]\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }[/tex]
where,
Diffusion rate of unknown gas = 155 mL/s
CO_2 diffusion rate = 102 mL/s
CO_2 molar mass = 44 g mol^-1
Unknown gas molercualr mass = M_unknown
Now placing these values to the above formula
[tex]\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}[/tex]
After solving this, the molecular mass of the unknown gas is
= 19.07 g mol^-1
For a single substance at atmospheric pressure, classify the following as describing a spontaneous process, a nonspontaneous process, or an equilibrium system.
A) Solid melting below its melting point
B) Gas condensing below its condensation point
C) Liquid vaporizing above its boiling point
D) Liquid freezing below its freezing point
E) Liquid freezing above its freezing point
F) Solid melting above its melting point
G) Liquid and gas together at boiling point with no net condensation or vaporization
H) Gas condensing above its condensation point
I) Solid and liquid together at the melting point with no net freezing or melting
Answer:
Spontaneous process- This is the process that occurs on its own without the application of any external energy or other factor. They include
B) Gas condensing below its condensation point
C) Liquid vaporizing above its boiling point
D) Liquid freezing below its freezing point
F) Solid melting above its melting point
Non spontaneous - This is the process that doesn’t occurs on its own and requires the application of any external energy or factor. They include
A) Solid melting below its melting point
E) Liquid freezing above its freezing point
H) Gas condensing above its condensation point
Equilibrium system
G) Liquid and gas together at boiling point with no net condensation or vaporization
I) Solid and liquid together at the melting point with no net freezing or melting
A) Solid melting below its melting point - nonspontaneous process
B) Gas condensing below its condensation point - spontaneous process
C) Liquid vaporizing above its boiling point - spontaneous process
D) Liquid freezing below its freezing point - spontaneous process
E) Liquid freezing above its freezing point - nonspontaneous process
F) Solid melting above its melting point - spontaneous process
G) Liquid and gas together at boiling point with no net condensation or vaporization - Equilibrium system
H) Gas condensing above its condensation point - nonspontaneous process
I) Solid and liquid together at the melting point with no net freezing or melting - Equilibrium system
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g Increasing the number of unsaturations in a fatty acid ____________ the melting temperature of the fatty acid.
Answer:
Decreases
Explanation:
Fatty acid which have the double bond or triple bond are called unsaturated fatty acids. Because of the double or triple bond, unsaturated fatty acids are loosely packed and form some distance among molecules which lowers the melting point of unsaturated fatty acids.
So, if the unsaturation of fatty acid will increase, it leads to more branched and loosely packed molecules and decreases the melting temperature accordingly.
g Air contains nitrogen, oxygen, argon, and trace gases. If the partial pressure of nitrogen is 592 mm Hg, oxygen is 160 mm Hg, argon is 7 mm Hg, and trace gas is 1 mm Hg, what is the atmospheric pressure
Answer:
Explanation:
Atmospheric pressure = partial pressure of nitrogen + partial pressure of oxygen + partial pressure of argon + partial pressure of trace element
putting the given values
Atmospheric pressure = 592 + 160 + 7 + 1
= 760 mm of Hg .
Balance this equation: __ UO2(s) + __ HF(ℓ) → __ UF4(s) + __ H2O(ℓ) Though you would not normally do so, enter the coefficient of "1" if needed. UO2(s) HF(ℓ) UF4(s) H2O(ℓ)
Answer:
The balanced equation is given below:
UO2(s) + 4HF(l) —› UF4(s) + 2H2O(l)
The coefficients are: 1, 4, 1, 2
Explanation:
UO2(s) + HF(l) —› UF4(s) + H2O(l)
The above equation can be balance as follow:
There are 2 atoms of O on the left side and 1 atom on the right side. It can be balanced by writing 2 in front of H2O as shown below:
UO2(s) + HF(l) —› UF4(s) + 2H2O(l)
There are 4 atoms of F on the right side and 1 atom on the left side. It can be balanced by writing 4 in front of HF as shown below:
UO2(s) + 4HF(l) —› UF4(s) + 2H2O(l)
Now, the equation is balanced.
The coefficients are: 1, 4, 1, 2.
A reaction has the rate law, rate = k[A]3[B]. Which change will cause the greatest increase in the reaction rate?
Answer:
Increasing the concentration of A will cause the greatest change over the rate.
Explanation:
Hello,
In this case, considering the given rate law, which is third-order with respect to A, changing its concentration, the rate will be significantly modified. For instance, suppose a concentration of A and B of 1M and a symbolic rate constant (k), this causes the rate to be:
[tex]r=k[1M]^3[1M]=1k\frac{M^4}{s}[/tex]
Then, if we change the concentration of A to 2 M holding the concentration of B in 1 M, the new rate constant will be:
[tex]r=k[2M]^3[1M]=8k\frac{M^4}{s}[/tex]
Nevertheless, if we hold the concentration of A in 1 M and the concentration of B is now 2 M (same change), the new rate constant is:
[tex]r=k[1M]^3[2M]=2k\frac{M^4}{s}[/tex]
It means that increasing the concentration of A will cause the greatest change over the rate.
Best regards.
The change in concentration of A will cause the greatest increase in the reaction rate.
Rate of reaction :Given reaction has rate law ,
[tex]rate=k[A]^{3} [B][/tex]
which is third-order with respect to concentration of A
From rate law equation, It is observed that the degree of concentration of A is three.
It means that small change in concentration of A result large change in rate of reaction.Thus, the change in concentration of A will cause the greatest increase in the reaction rate.Learn more about the rate of reaction here :
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A calibration curve constructed from absorbance values of solutions containing a known concentration of permanganate ions has the following best-fit line:
y = (3.62× 10^3 L/mol) x
where y is the absorbance of the solution at 525 nm and x is the concentration of MnO4- (aq) in mol/L. The path length of the cuvettes used in the experiment is 1 cm. Based on this information, what is the molar absorptivity of MnO4- (aq) at 525 nm?
Answer:
3.62×10³ L/mol
Explanation:
Beer-Lambert law relates the absorbance of a sample and its concentration. Its formula is:
A = ε×C×l
Where A is absorbance of the sample, ε is molar absorptivity (A constant f each sample), C its concentration and l is path length
Now, the formula obtained was:
y = (3.62×10³ L/mol) x
Where Y ia absorbance = A, x its concentration = C and 1cm is path length.
You can write:
A = (3.62×10³ L/mol)×C×l
That means, molar absorptivity of your sample under the meaured conditions is:
3.62×10³ L/mol
Which is one characteristic of producers?
They recycle nutrients.
They do not eat other organisms.
They break down waste for energy.
They use other organisms for energy.
Answer: They use other organisms for energy
Explanation:
Answer:
they use other organisms for energy
Explanation: d
A glass cylinder contains 2 gases at a pressure of 106 kPa. If one gas is at 7 kPa, what is the pressure of attributed to the other gas? a) 9 kPa b) 99 kPa c) 113 kPa d) 7 kPa e) 2 kPa (URGENT)
Answer:
b) 99 kPa
Explanation:
According to Daltons law of partial pressure, the total pressure of a mixture of two or more non reactive gases is the sum of their individual pressures. Let the total pressure of a mixture of n number of gases be [tex]P_{total}[/tex] and their individual pressure be [tex]P_1,P_2,P_3,\ .\ .\ .\ ,\ P_n[/tex], According to Daltons partial pressure law:
[tex]P_{total}=P_1+P_2+P_3+.\ .\ .+P_n[/tex]
Since A glass cylinder contains 2 gases at a pressure of 106 kPa, therefore n = 2. Also one gas ([tex]P_1[/tex]) is at 7 kPa. Using Daltons partial pressure law:
[tex]P_{total}=P_1+P_2+P_3+.\ .\ .+P_n\\P_{total}=P_1+P_2\\106\ kPa=7\ kPa+P_2\\P_2=106\ kPa-7\ kPa\\P_2=99\ kPa[/tex]
According to the collision theory, all collisions do not lead to reaction. Which choice gives both reasons why not all collisions between reactant molecules lead to reaction? 1. The total energy of two colliding molecules is less than some minimum amount of energy. 2. Molecules cannot react with each other unless a catalyst is present. 3. Molecules that are improperly oriented during collision will not react. 4. Solids cannot react with gases.
Answer: 3. Molecules that are improperly oriented during collision will not react.
Explanation:
According to the collision theory , the number of collisions that take place per unit volume of the reaction mixture is called collision frequency. The effective collisions are ones which result into the formation of products.
It depends on two factors:-
1. Energy factor: For collision to be effective, the colliding molecules must have energy more than a particular value called as threshold energy.
2. Orientation factor: Also the colliding molecules must have proper orientation at the time of collision to result into formation of products.
Thus not all collisions between reactant molecules lead to reaction because molecules that are improperly oriented during collision will not react.
BEARINGS
Question 12 (SSCE 1994 May/June)
(a) A village P is 10km from a lorry station, Q, on
a bearing 065º. Another village R, is 8km from
Q on a bearing 155º. Calculate
(i) the distance of R from P to the nearest
kilometer
(ii) the bearing of R from P to the nearest degree
(b) M is a village on PR such that QM is
perpendicular to PR. Find the distance of M
from P to the nearest kilometer.
Answer:
a. (i) the distance of R from P is 13 Km to the nearest kilometer
(ii) the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree
b. the distance of m from P is 11 Km to the nearest kilometre
Explanation:
a) A triangle PQR is formed. Q = 90°, p = 8 km; r = 10 km; distance of R from P is q is to be found.
(i) Using the cosine rule: q² = p² + r² - 2prCosQ
q² = 8² + 10² - 2 * 8 * 10 * Cos90
q² = 64 + 100 + 0
q² = 164
q = 13 Km to the nearest kilometre
Therefore, the distance of R from P is 13 Km to the nearest kilometer
(ii) the bearing of R from P
The angle at P is found using the formula Cos P = (q² + r² - p²)/2qr
Cos P = 13² + 10² - 8²/2 * 13 *10
Cos P = 0.7884
P = Cos⁻¹ 0.7884
P = 38°
Therefore, the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree
Note : 25° is alternate (Northwest) to 65°at P
b) A right-angled triangle QMP is formed
using the trigonometrical ratios; cos Θ = adjacent/hypotenuse
where the hypotenuse side = 10 km, adjacent side = distance of M from P, x
cos P = x/10
x = cos 38 * 10
x = 11 Km to the nearest kilometre
Therefore, the distance of m from P is 11 Km to the nearest kilometre
An amphoteric salt is one that contains an anion that can act as either an acid or a base in water. Baking soda, NaHCO3, is an example. By combining the ionization and hydrolysis reactions of the anion, you get the principle reaction that occurs when this salt is dissolved in water.
2HCO3-(aq) CO32-(aq) + H2CO3(aq)
The pH for such, a solution is given by
pH = pKa1 + pKa2/2
where Ka1 and Ka2 are the ionization constants of the acid (in this case, H2CO3). Note that the pH of the solution is independent of the salt concentration. Calculate the pH of a NaHCO3 solution.
Answer:
pH = 8.34
Explanation:
The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:
H₂CO₃ ⇄ HCO₃⁻ + H⁺ Ka1 -Here, HCO₃⁻ is acting as a base-
HCO₃⁻⇄ CO₃²⁻ + H⁺ Ka2 -Here, is acting as an acid-
Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:
pKa1 = 6.37; pKa2 = 10.32
As the pH of amphoteric salts is:
pH = (pKa1 + pKa2) / 2
pH = 8.34Which of the following is an alkaline earth metal?
A. Silicon (Si)
B. Magnesium (Mg)
C. Carbon (C)
D. Aluminum (AI)
Answer:
B
Explanation:
The alkaline earth metals are the elements located in Group 2. The only element out of our choices that is in Group 2 is Magnesium.
3. There are many different primary standards that could be used in a standardization titration. What are the criteria for a primary standard
Answer:
High purity.
Stability (low reactivity)
Low hygroscopicity (to minimize weight changes due to humidity)
Explanation:
There are different primary standards that could be used in a standardization titration in order to achieve the best and accurate result possible. These standards include high purity,stability and low hygoscropicity .
A high purity means the reactants lack impurities which could affect the result. Stability also ensures that there is non reactivity with elements/compounds in the atmosphere while low hygroscopicity ensures weight changes are minimized due to humidity.
Identify the energy transformations in the following actions.
a. Turning on a space heater
b. Dropping an apple core into the garbage
c. Climbing up a rope ladder
d. Starting a car
e. Turning on a flashlight
Think about each action as a before and after. For instance, before the space heater is turned on, what kind of energy is there? After the space heater is turned on, what happens? This change from before to after will help you identify the two types of energy involved.
Identify five more types of energy transformations that you see at home, at school, or outdoors. Make sure to name the action, such as turning on a light, as well as the two types of energy involved. Remember, for energy to be transformed, the type of energy before and after a task must be different.
Answer:
to. Turn on a heater (electrical energy that is transformed into heat energy)
yes. Throwing an apple core in the trash (chemical energy that is transformed into nutritional energy for decomposers)
C. Climbing a rope ladder (chemical energy from our food into mechanical energy that allows us to climb the ladder)
re. Starting a car (electrochemical energy in mechanical energy)
me. Turning on a flashlight (chemical energy from the battery into light energy)
Five daily actions that exemplify the transformation of energy:
Do physical activity (change of metabolic energy in mechanics)
Turn on a heater (electrical energy that is transformed into heat energy)
Lighting a stove (Chemical energy product of combustion that results in the transformation of that energy into heat)
Turn on a cell phone (chemical energy characteristic of the battery that is transformed into sound and light energy)
Riding a bicycle (nutritional energy or energy of the metabolism that is transformed into mechanical energy)
Explanation:
The energy is never lost, it is always transformed.
It is one of the great principles of physics and is the reason why the universe is infinite.
Identify the elements that have the following abbreviated electron configurations.
A) [Ne] 3s23p5.
B) [Ar] 4s23d7.
C) [Xe] 6s1.
Answer:
A) Chlorine (Cl)
B) Cobalt (Co)
C) Caesium (Cs)
Hope this helps.
The abbreviated electron configurations that was given in the question belongs to
Chlorine (Cl)
Cobalt (Co)
Caesium (Cs) respectively.
Electronic configurations can be regarded as the electronic structure, which is the way an electrons is arranged in energy levels towards an atomic nucleus.The electron configurations is very useful when describing the orbitals of an atom in its ground state.To calculate an electron configuration, we can put the periodic table into sections, and this section will represent the atomic orbitals which is the regions that house the electrons. Groups one of the period table and two belongs to s-block, group 3 through 12 belongs to the d-block, while 13 to 18 can be attributed to p-block ,The rows that is found at bottom are the f-blockTherefore, electron configurations explain orbitals of an atom when it is in it's ground state.
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When electrons are filling energy levels, the lowest energy sublevels are occupied first. This is ____________.
When electrons are filling energy levels, the lowest energy sublevels are occupied first. This is Hund's rule.
Hund's rules state that:
Every orbital in a sublevel has to be singularly occupied before any other orbital is able to be doubly occupied.
All of the electrons in single occupied orbitals have to have the same spin to maximize the total spin.
Which Carbon is the triple bound attached to in 6-ethyl-2-octyne?
-first
-fourth
-third
-second
Answer:
-second
Explanation:
6-ethyl-2-octyne is an unsaturated compound with a triple bond.
6-ethyl-2-octyne will have a triple bound attached to the second carbon. The suffix -yne suggests that compound carry a triple bond and the number "2" before suffix refers to the position of triple bond that is second carbon.
Hence, the correct option is "-second ".
Calculate the molality of a solution prepared by dissolved 19.9 g of kcl in 750ml of water
First, we find the molar mass of KCl which is AK+ACl=39+35.5=74.5g/mole
now we find out the number of moles by dividing the given mass to the molar mass n=m/M=19.9/74.5=0.26 moles. The molarity of a solution is equal to the number of moles divided by the volume of the solution. =0.26moles/0.75liters=0.346M
write the IUPAC name OF THE FOLLOWING COMPOUNDS
Answer:
Explanation:
a) 2 chloro butane
b) 2-3 dimethyl butane
c) 2 bromo 3 nitro pentane
d) 2-3 trimethyl pentane
e) 2-bromo,3-methyl,4-nitro hexane
f) 2-methyl cyclo butane
How many molecules of CaCl2 are equivalent to 75.9g CaCl2 (Ca=40.08g/mol, CL=35.45g/mol)
Answer:
[tex]\large \boxed{4.12 \times 10^{23}\text{ formula unis of CaCl}_{2}}$}[/tex]
Explanation:
You must calculate the moles of CaCl₂, then convert to formula units of CaCl₂.
1. Molar mass of CaCl₂
CaCl₂ = 40.08 + 2×35.45 = 40.08 + 70.90 = 110.98 g/mol
2. Moles of CaCl₂ [tex]\text{Moles of CaCl}_{2} = \text{75.9 g CaCl}_{2} \times \dfrac{\text{1 mol CaCl}_{2}}{\text{110.98 g CaCl}_{2}} = \text{0.6839 mol CaCl}_{2}[/tex]
3. Formula units of CaCl₂
[tex]\text{No. of formula units} = \text{0.6839 mol CaCl}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules CaCl}_{2}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{4.12 \times 10^{23}}\textbf{ formula units CaCl}_{2}\\\text{There are $\large \boxed{\mathbf{4.12 \times 10^{23}}\textbf{ formula units of CaCl}_{2}}$}[/tex]
Convert the following measurement
Answer:
6.9 Kg/mol•dL
Explanation:
To convert 6.9×10⁴ g/mol•L to kg/mol•dL,
First, we shall convert to kg/mol•L.
This can be achieved by doing the following:
Recall: 1 g = 1×10¯³ Kg
1 g/mol•L = 1×10¯³ Kg/mol•L.
Therefore,
6.9×10⁴ g/mol•L = 6.9×10⁴× 1×10¯³
6.9×10⁴ g/mol•L = 69 Kg/mol•L
Finally, we shall convert 69 Kg/mol•L to Kg/mol•dL.
This is illustrated below:
Recall: 1 L = 10 dL
1 Kg/mol•L = 1×10¯¹ Kg/mol•dL
Therefore,
69 Kg/mol•L = 69 × 1×10¯¹
69 Kg/mol•L = 6.9 Kg/mol•dL
Therefore, 6.9×10⁴ g/mol•L is equivalent to 6.9 Kg/mol•dL.
Charcoal from the dwelling level of the Lascaux Cave in France gives an average count of 0.97 disintegrations of ^14 C per minute per gram of sample. Living wood gives 6.68 disintegrations per minute per gram. Estimate the date of occupation and hence the probable date of the wall painting in the Lascaux Cave. Hint: Disintegrations per minute per gram" has the same units as the time-derivative of concentration for a radioactive decay model. (You may use the fact that the half-life of ^14C is 5568 years.)
Answer:
Explanation:
count given by old sample = .97 disintegrations per minute per gram
count given by fresh sample = 6.68 disintegrations per minute per gram
Half life of radioactive carbon = 5568 years
rate of disintegration
dN / dt = λ N
In other words rate of disintegration is proportional to no of radioactive atoms present . As number reduces rate also reduces .
Let initial no of radioactive be N₀ and after time t , number reduces to N
N₀ / N = 6.68 / .97
Now
[tex]N=N_0e^{-\lambda t}[/tex]
[tex]\frac{N}{N_0} =e^{-\lambda t}[/tex]
[tex]\frac{6.68}{.97} = e^{\lambda t}[/tex]
λ is disintegration constant
λ = .693 / half life
= .693 / 5568
= .00012446 year⁻¹
Putting the values in the equation above
[tex]\frac{6.68}{.97} = e^{.00012446\times t}[/tex]
[tex]6.8866 = e^{.00012446\times t}[/tex]
1.929577 = .00012446 t
t = 15503.6 years .
An atom with 19 protons and 18 neutrons is a(n)
A. Isotope of potassium(K)
B. Standard atom of argon(Ar)
C. Standard atom of (K)
D. Isotope of argon (Ar)
Answer:
A
Explanation:
The number of protons indicates the element so we know it's potassium. To get the number of neutrons you subtract the number of protons (19) from the mass number which for potassium is 39.
39-19=20 neutrons
Because you have 18 neutrons then yours would be an isotope.
Answer: A. Isotope of potassium(K)
Explanation: Founders Educere answer.
Select the correct answer from each drop-down menu.
Pictures that we receive from space are of the
✓ because it takes time for
to reach Earth.
Answer:
Pictures that we receive from space are of the past because it takes time for light to reach Earth.
Explanation:
For example, Mars is so far away that, depending on its position in orbit, a picture from Mars takes between 4 min and 24 min to reach Earth.
Answer: Pictures that we receive from space are of the
past
because it takes time for
light
to reach Earth.
Explanation:
Which of the following sets of quantum numbers is NOT allowed? a. n = 5, l= 4, ml= –2, ms = +1/2 b. n = 2, l = 1, ml= 0, ms = +1/2 c. n = 4, l= 4, ml= 0, ms = –1/2 d. n = 3, l= 2, ml= –1, ms = +1/2 e. n = 2, l= 0, ml= 0, ms = -1/2
Answer:
C
Explanation:
the n value must always be greater than the l value
Out of the following set of quantum numbers ,set C is not allowed as the azimuthal and principal quantum numbers are same.
What are quantum numbers?Quantum numbers are the numbers which describe the values of conserved quantities with respect to the dynamics of a quantum system.They correspond to the Eigen values of operators which commute with the Hamiltonian quantities.
The Hamiltonian quantities can be known with precision simultaneously with the system's energy.Quantum numbers can take values of discrete sets of integers or even half-integers even though they can approach infinity in some cases.
They can specifically describe energy levels of electrons, and can also explain angular momentum,spin,etc.These are used to describe the path of an electron in an atom ,when the quantum numbers of all atoms are combined they must comply with the Schrodinger equation.
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2
22. A sodium chloride solution is 15.0% m/m%. Calculate mass of sodium chloride in 219 g solution.
14.2g
80.38
11.2 g
32.9 g
Answer: The mass of sodium chloride in 219 g solution is 32.9 g
Explanation:
To calculate the mass percent of element in a given compound, we use the formula:
[tex]\text{Mass percent of A}=\frac{\text{Mass of A}}{\text{mass of A +mass of B}}\times 100[/tex]
To find mass of sodium chloride in solution:
[tex]\text{Mass percent of sodium chloride}=\frac{\text{Mass of sodium chloride}}{\text{mass of solution}}\times 100[/tex]
Mass percent of sodium chloride= 15.0 %
Mass of solution = 219g
[tex]15=\frac{\text{Mass of sodium chloride}}{219}\times 100[/tex]
[tex]{\text{Mass of sodium chloride}=32.9g[/tex]
Thus mass of sodium chloride in 219 g solution is 32.9 g
Generally, systems move spontaneously in the direction of increasing entropy. TRUE FALSE
Answer:
true
Explanation:
How long should you hold the iron on the hair to heat the strand and set the base ?
A) 5 seconds
B) 15 seconds
C) 30 seconds
D) 1 minute
Determine the volume occupied by 10 mol of helium at 27 ° C and 82 atm
please.
Answer:
3.00 L
Explanation:
Convert the pressure to Pascals.
P = 82 atm × (101325 Pa/atm)
P = 8,308,650 Pa
Convert temperature to Kelvins.
T = 27°C + 273
T = 300 K
Use ideal gas law:
PV = nRT
(8,308,650 Pa) V = (10 mol) (8.314 J/mol/K) (300 K)
V = 0.00300 m³
If desired, convert to liters.
V = (0.00300 m³) (1000 L/m³)
V = 3.00 L
Answer:
[tex]\large \boxed{\text{3.0 L}}[/tex]
Explanation:
[tex]\begin{array}{rcl}pV &=& nRT\\\text{82 atm} \times V & = & \text{10 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{300.15 K}\\82V & = & \text{246 L}\\V & = & \textbf{3.0 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{3.0 L}}$}[/tex]